05/07/2021, 09:27 AM
Ok, I'm really confused now. I think there is a problem and maybe I know where it lies.
We have to reach a consensus on the definitions. Otherwise we will not be able to turn your claims into fully-formal proofs anytime soon.
I just make some points that makes the absence of perfect consensus evident.
-Yes, a function can not be multivalued and bijective at the same time.
-I say bijective because it is the only case where I can see you can prove that those sets are actually equal (when non-empty). In fact when we leave the bijectivity, the inverses are not function but multivalued objects. In that case, I claim, those sets are much richer and inclusion or equality of the sets can fail badly. I see that the way you state the first part of Law 5 is too strong to hold, in any interpretation. It can not be satisfied because composing on the left by a generalized iteration could shrink the set. Doing that on the right could shrink the set in a different manner. So I exclude that you can come up with an equality of the three sets in general.
-Mutivalued functions ARE perfectly defined algebraically. Maybe are hard to compute or to manage in your computations but are not really a contradiction.
-The generalization of odd function reduces to satisfying hf=fh. If f conjugates the function h(x)=-x then, by definition, f is odd.
-Let's call f generalized h-even function if fh=f. This means that f is in the set \( [h,{\rm id}] \). Now I can see the for non-zero natual numer of iterations of f we get an h-even function (easy to prove):
\( \forall n\in{\mathbb N}_0.f\in [h,{\rm id}]\Rightarrow f^n\in [h,{\rm id}] \)
Being h-even is equivalent to not being injective so I agree that if f is h-even function the inverse is not a single valued function.
You claim that if we consider the multivalued case we can somehow recover the previous implication when n is a non-zero integer, e.g -3, -2, -1, 1, 2 ... and so on. Ok I can believe that. What I can't see is how you can extend this to generalized iterations of f(when they exists).
In the end, I invite you to make formal the definitions.
Given a set \( X \) and two functions \( f,g:X\to X \) I define the set
\( [f,g]:=\{\chi:X\to X \,\|\, \forall x\in X. \chi(f(x))=g(\chi(x)) \} \)
Let me try to formalize what you are talking about. I hope JmsNx can also help me here. You seem to define your set as follows:
Given a set \( X \), assume \( X={\mathbb C} \),
given subsets \( F_0\subseteq{\mathbb C} \), \( F_1\subseteq{\mathbb C} \), \( G_0\subseteq{\mathbb C} \), \( G_1\subseteq{\mathbb C} \) such that \( F_0\cap F_1\neq {\emptyset} \) and \( G_0\cap G_1\neq \emptyset \) are non-empty.
Given two possible mutivalud functions \( f: F_0\to F_1 \) and \( g:G_0\to G_1 \) (to be really precise we should write \( f\subset F_0\times F_1 \) and \( g\subset G_0\times G_1 \) )
Define the set
\( \{f;g\}:=\{\zeta: F_0\cap F_1\to {\mathbb C} \,\|\, \exists D\subseteq F_0\cap F_1 \forall z\in D. \zeta(f(z))=g(\zeta(z)) \} \)
If this is the case, then the theory of those sets is, I suspect, a billion times more ill behaved and weird than theory of my sets is. I limit myself only with functions well defined on given domains. You consider freely "bundles" of multivalued functions with non-specified domains. It is indeed a more flexible setting for sure. The only way to be more flexible than this is considering non-functional relations.
I'm afraid that turning this mess into category theory is beyond my powers atm. It is in fact the realm of differential geometry and sheaf theory. I'm in deep water.
ps: I'm not a moderator but I suggest you to remove from your answers the whole citation of my message when it is too lengthy or when it is clear who are you writing to or to what are you referring. We can make the thread more readable and short (less pages).
We have to reach a consensus on the definitions. Otherwise we will not be able to turn your claims into fully-formal proofs anytime soon.
I just make some points that makes the absence of perfect consensus evident.
-Yes, a function can not be multivalued and bijective at the same time.
-I say bijective because it is the only case where I can see you can prove that those sets are actually equal (when non-empty). In fact when we leave the bijectivity, the inverses are not function but multivalued objects. In that case, I claim, those sets are much richer and inclusion or equality of the sets can fail badly. I see that the way you state the first part of Law 5 is too strong to hold, in any interpretation. It can not be satisfied because composing on the left by a generalized iteration could shrink the set. Doing that on the right could shrink the set in a different manner. So I exclude that you can come up with an equality of the three sets in general.
-Mutivalued functions ARE perfectly defined algebraically. Maybe are hard to compute or to manage in your computations but are not really a contradiction.
-The generalization of odd function reduces to satisfying hf=fh. If f conjugates the function h(x)=-x then, by definition, f is odd.
-Let's call f generalized h-even function if fh=f. This means that f is in the set \( [h,{\rm id}] \). Now I can see the for non-zero natual numer of iterations of f we get an h-even function (easy to prove):
\( \forall n\in{\mathbb N}_0.f\in [h,{\rm id}]\Rightarrow f^n\in [h,{\rm id}] \)
Being h-even is equivalent to not being injective so I agree that if f is h-even function the inverse is not a single valued function.
You claim that if we consider the multivalued case we can somehow recover the previous implication when n is a non-zero integer, e.g -3, -2, -1, 1, 2 ... and so on. Ok I can believe that. What I can't see is how you can extend this to generalized iterations of f(when they exists).
In the end, I invite you to make formal the definitions.
Given a set \( X \) and two functions \( f,g:X\to X \) I define the set
\( [f,g]:=\{\chi:X\to X \,\|\, \forall x\in X. \chi(f(x))=g(\chi(x)) \} \)
Let me try to formalize what you are talking about. I hope JmsNx can also help me here. You seem to define your set as follows:
Given a set \( X \), assume \( X={\mathbb C} \),
given subsets \( F_0\subseteq{\mathbb C} \), \( F_1\subseteq{\mathbb C} \), \( G_0\subseteq{\mathbb C} \), \( G_1\subseteq{\mathbb C} \) such that \( F_0\cap F_1\neq {\emptyset} \) and \( G_0\cap G_1\neq \emptyset \) are non-empty.
Given two possible mutivalud functions \( f: F_0\to F_1 \) and \( g:G_0\to G_1 \) (to be really precise we should write \( f\subset F_0\times F_1 \) and \( g\subset G_0\times G_1 \) )
Define the set
\( \{f;g\}:=\{\zeta: F_0\cap F_1\to {\mathbb C} \,\|\, \exists D\subseteq F_0\cap F_1 \forall z\in D. \zeta(f(z))=g(\zeta(z)) \} \)
If this is the case, then the theory of those sets is, I suspect, a billion times more ill behaved and weird than theory of my sets is. I limit myself only with functions well defined on given domains. You consider freely "bundles" of multivalued functions with non-specified domains. It is indeed a more flexible setting for sure. The only way to be more flexible than this is considering non-functional relations.
I'm afraid that turning this mess into category theory is beyond my powers atm. It is in fact the realm of differential geometry and sheaf theory. I'm in deep water.
ps: I'm not a moderator but I suggest you to remove from your answers the whole citation of my message when it is too lengthy or when it is clear who are you writing to or to what are you referring. We can make the thread more readable and short (less pages).
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)